Let $f : \mathbb{R} \to \mathbb{R}$ be approximated arbitrarily well by polynomials of bounded degree. Prove $f$ is a polynomial.

Question

I am trying to prove that if a function $f : \mathbb{R} \to \mathbb{R}$ can be approximated arbitrarily well by polynomials of bounded degree, then $f$ itself must be a polynomial.

For starters, let $\{ g_m \mid g_m : \mathbb{R} \to \mathbb{R}\}_{m=1}^\infty$ be a sequence of polynomials with degree at most $d$ which uniformly converges to $f$. And let the coefficients of the $m$'th approximator be $a_{mn}$ so that $$ g_m(x) = \sum_{n=0}^d a_{mn} \, x^n. $$

By "uniform convergence", I mean that given $\epsilon > 0$, there exists an $M_\epsilon$ such that for all $m \ge M_\epsilon$ we have $$ \sup_{x \in \mathbb{R}} |f(x) - g_m(x)| < \epsilon. $$

Clearly, proving the claim is equivalent to proving that for each $n$ between 0 and $d$, the sequence of coefficients $a_{1n}, a_{2n}, \ldots$ converges to a some limiting value $a_n^*$.

However, it's not clear to me how to prove this statement. Any hints or guidance is greatly appreciated.

Answer 1

For each $n$ you can find some $$P_n(x)=a_d^nx^d+...+a_1^nx+a_0^n$$ such that $$\sup_{x \in \mathbb{R}} |f(x) - P_n(x)| < \frac{1}{n}$$

Then, for each $n,m$ you have $$\sup_{x \in \mathbb{R}} |P_m(x) - P_n(x)| \leq \sup_{x \in \mathbb{R}} |f(x) - P_n(x)| +\sup_{x \in \mathbb{R}} |f(x) - P_m(x)| < \frac{1}{n}+ \frac{1}{m}$$

This shows that $P_m(x)-P_n(x)$ is a polynomial which is bounded on $\mathbb R$ and hence a constant. Therefore, there exists some polynomial $P(x)$ and some $c_n \in \mathbb R$ such that $$P_n(x)=P(x)+c_n$$

By the above you get $$\left| c_n -c_m \right| =\sup_{x \in \mathbb{R}} |P_m(x) - P_n(x)| < \frac{1}{n}+ \frac{1}{m} $$

Therefore, $c_n$ is Cauchy and hence convergent to some $c$.

We claim that $f(x)=P(x)+c$.

Let $\epsilon >0$. Pick some $N$ such that, for all $n >N$ we have $$\frac{1}{n} < \frac{\epsilon}{2}\\ |c_n-c|< \frac{\epsilon}{2}$$

Let $n >N$ be fixed but arbitrary. Then $$\sup_{x \in \mathbb{R}} |f(x) - P(x)| \leq \sup_{x \in \mathbb{R}} |f(x) - P_n(x)| +\sup_{x \in \mathbb{R}} |P_n(x) - P(x)| < \frac{1}{n}+ |c_n-c| <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

This shows that $$\sup_{x \in \mathbb{R}} |f(x) - P(x)| < \epsilon$$ for all $\epsilon >0$.

Answer 2

First pick a polynomial $g(x)$ for $\varepsilon = 1$, i.e., $\sup\limits_{x\in \mathbb{R}}|f(x)-g(x)|< 1$. Let $h(x)= f(x)-g(x)$. Then the same condition holds for $h(x)$: To obtain an approximation with error $\varepsilon$, find one for $f(x)$ and subtract $g(x)$.

So it is enough to verify the assertion for bounded functions. But that is trivial: a bounded function with this property must be constant. Indirect proof: if it had two different values (say at points $x,y$), let $\delta=|h(x)-h(y)|$, and put $\varepsilon:=\delta/3$. The approximating polynomial must be constant (otherwise we have a problem with the limit at infinity), but no constant is close enough at both $x$ and $y$. Hence, $h(x)$ is a constant, and then $f(x)= g(x)+h(x)$ is a polynomial.

Answer 3

Here is a different approach: Denote $\mathcal{B} [-t,t]$ the space of bounded functions on the interval $[-t,t]$. We consider this space with the supremum norm $\Vert . \Vert _\infty $, which makes it a normed space. Denote the space of polynomials with degree at most $d$ by $W=\mathbb{R}_{\leq d} [x] $, these are continues thus bounded function on the interval $[-t,t]$. But we know more than that: W is a finite dimensional subspace of $\mathcal{B} [-t,t]$ for every t, thus it's closed.

If you can approximate $f:\mathbb{R} \rightarrow \mathbb{R}$ aritrarily well on $\mathbb{R}$ by elements in $W$, of course you can approximate it arbitrarily well in $\mathcal{B} [-t,t]$ for every $t>0$ (and $f \in \mathcal{B} [-t,t]$). But $W$ is closed and thus $f \vert _{[-t,t]} \in W$ for all $t$, i.e. it is a polynomial on every $[-t,t]$. But every two polynomials that agree on an open set must be the same polynomial (they have the same derivatives, which give the coefficients, i.e. Taylor Expansion), thus we see $f$ is a polynomial on the whole of $\mathbb{R}$

Answer 4

Since $f$ is a uniform limit of polynomials (contiunous functions), in particular it is continuous.

Consider the space $C(\mathbb{R})$ of continuous functions $\mathbb{R} \to \mathbb{R}$ and equip it with a family of seminorms $\|\cdot\|_{\infty, [a,b]}$ given by $$\|g\|_{\infty, [a,b]} = \sup_{x\in[a,b]}|g(x)|$$

for all segments $[a,b] \subseteq \mathbb{R}$.

This turns $C(\mathbb{R})$ into a locally convex topological vector space.

Assume that $f$ can be uniformly approximated by polynomials $(p_k)_k$ of degree $\le n$.

The subspace $\mathbb{R}_{\le n}[x]$ of real polynomials of degree $\le n$ is a finite dimensional subspace of $C(\mathbb{R})$ and hence it is closed in $C(\mathbb{R})$. Since $p_k \to f$ uniformly on $\mathbb{R}$, in particular $\|f -p_k\|_{\infty, [a,b]} \to 0$ for all segments $[a,b] \subseteq \mathbb{R}$.

Hence $p_k \to f$ in the above topology so $f$ is in the closure of $\mathbb{R}_{\le n}[x]$. Since the subspace is closed, we conclude $f \in \mathbb{R}_{\le n}[x]$.